數學問題(急)要步驟

︳︳x-2︳-1︳= a
︳x-2︳-1= a or ︳x-2︳-1= -a
︳x-2︳=1+a or ︳x-2︳=1-a
x= a+3 x=3-a
or
x=1-a x=a+1

注意 a+3 =/= a+1 及 1-a =/= 3-a
已知方程︳︳x-2︳-1︳= a有四個不同的解
即
a+3 =/= 1-a 1-a =/= a+1 3-a=/= a+1

a =/=-1, a=/=0 a=/=1

2008-07-30 14:01:53 補充：
the method of squaring make you missing some solution

︳︳x-2︳-1︳= a
︳x-2︳-1=a or ︳x-2︳-1=-a
︳x-2︳=a+1 or ︳x-2︳=1-a
︳x-2︳^2=(a+1)^2=︳x-2︳^2=(1-a)^2
x^2-4x+4=a^2+2a+1or x^2-4x+4=1-2a+a^2
x^2-4x-(a^2+2a-3)=0 orx^2-4x-(a^2-2a-3)=0
方程︳︳x-2︳-1︳= a有四個不同的解
即每個方程都要有2個不同解
delta>0
4^2-4(-(a^2+2a-3))>0 or 4^2-4(-(a^2-2a-3))>0
16+4a^2+8a-12>0 or 16+4a^2-8a-12>0
a^2+2a+1>0 or a^2-2a+1>0
(a+1)^2>0 or (a-1)^2>0
a<-1 or a>-1 or a<1 or a>1
a的取值範圍是所有實數,除1,-1以外